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You graphed linear and exponential functions. Analyze the characteristics of graphs of quadratic functions. Graph quadratic functions. Then/Now

Concept

Graph these ordered pairs and connect them with a smooth curve. Graph a Parabola Use a table of values to graph y = x2 – x – 2. State the domain and range. Graph these ordered pairs and connect them with a smooth curve. Answer: domain: all real numbers; Example 1

Use a table of values to graph y = x2 + 2x + 3. C. D. Example 1

Identify Characteristics from Graphs A. Find the vertex, the equation of the axis of symmetry, and y-intercept of the graph. Step 1 Find the vertex. Because the parabola opens up, the vertex is located at the minimum point of the parabola. It is located at (2, –2). Example 2

Step 2 Find the axis of symmetry. Identify Characteristics from Graphs A. Find the vertex, the equation of the axis of symmetry, and y-intercept of the graph. Step 2 Find the axis of symmetry. The axis of symmetry is the line that goes through the vertex and divides the parabola into congruent halves. It is located at x = 2. Example 2

Step 3 Find the y-intercept. Identify Characteristics from Graphs A. Find the vertex, the equation of the axis of symmetry, and y-intercept of the graph. Step 3 Find the y-intercept. The y-intercept is the point where the graph intersects the y-axis. It is located at (0, 2), so the y-intercept is 2. Example 2

axis of symmetry: x = 2; y-intercept: 2 Identify Characteristics from Graphs A. Find the vertex, the equation of the axis of symmetry, and y-intercept of the graph. Answer: vertex: (2, –2); axis of symmetry: x = 2; y-intercept: 2 Example 2

Identify Characteristics from Graphs B. Find the vertex, the equation of the axis of symmetry, and y-intercept of the graph. Step 1 Find the vertex. The parabola opens down, so the vertex is located at the maximum point (2, 4). Example 2

Step 2 Find the axis of symmetry. Identify Characteristics from Graphs Step 2 Find the axis of symmetry. The axis of symmetry is located at x = 2. Step 3 Find the y-intercept. The y-intercept is where the parabola intersects the y-axis. it is located at (0, –4), so the y-intercept is –4. Answer: vertex: (2, 4); axis of symmetry: x = 2; y-intercept: –4 Example 2

A. Consider the graph of y = 3x2 – 6x + 1 A. Consider the graph of y = 3x2 – 6x + 1. Find the coordinates of the vertex. A. (–1, 10) B. (1, –2) C. (0, 1) D. (–1, –8) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 Example 2

B. Consider the graph of y = 3x2 – 6x + 1 B. Consider the graph of y = 3x2 – 6x + 1. Write the equation of the axis of symmetry. A. x = –6 B. x = 6 C. x = –1 D. x = 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 Example 2

Formula for the equation of the axis of symmetry Identify Characteristics from Functions A. Find the vertex, the equation of the axis of symmetry, and y-intercept of y = –2x2 – 8x – 2. Formula for the equation of the axis of symmetry a = –2, b = –8 Simplify. Example 3

The equation for the axis of symmetry is x = –2. Identify Characteristics from Functions The equation for the axis of symmetry is x = –2. To find the vertex, use the value you found for the axis of symmetry as the x-coordinate of the vertex. To find the y-coordinate, substitute that value for x in the original equation y = –2x2 – 8x – 2 Original equation = –2(–2)2 – 8(–2) – 2 x = –2 = 6 Simplify. The vertex is at (–2, 6). The y-intercept occurs at (0, c). So, the y-intercept is –2. Example 3

Answer: vertex: (–2, 6); axis of symmetry: x = –2; y-intercept: –2 Identify Characteristics from Functions A. Find the vertex, the equation of the axis of symmetry, and y-intercept of y = –2x2 – 8x – 2. Answer: vertex: (–2, 6); axis of symmetry: x = –2; y-intercept: –2 Example 3

Formula for the equation of the axis of symmetry Identify Characteristics from Functions B. Find the vertex, the equation of the axis of symmetry, and y-intercept of y = 3x2 + 6x – 2. Formula for the equation of the axis of symmetry a = 3, b = 6 Simplify. Example 3

The equation for the axis of symmetry is x = –1. Identify Characteristics from Functions The equation for the axis of symmetry is x = –1. To find the vertex, use the value you found for the axis of symmetry as the x-coordinate of the vertex. To find the y-coordinate, substitute that value for x in the original equation. y = 3x2 + 6x – 2 Original equation = 3(–1)2 + 6(–1) – 2 x = –1 = –5 Simplify. The vertex is at (–1, –5). The y-intercept occurs at (0, c). So, the y-intercept is –2. Example 3

Answer: vertex: (–1, –5); axis of symmetry: x = –1; y-intercept: –2 Identify Characteristics from Functions B. Find the vertex, the equation of the axis of symmetry, and y-intercept of y = 3x2 + 6x – 2. Answer: vertex: (–1, –5); axis of symmetry: x = –1; y-intercept: –2 Example 3

A. Find the vertex for y = x2 + 2x – 3. B. (1, –2) C. (–1, –4) D. (–2, –3) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 Example 3

B. Find the equation of the axis of symmetry for y = 7x2 – 7x – 5. B. x = 1.5 C. x = 1 D. x = –7 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 Example 3

Assignment:1/6/16 Page 549 – 550 Problems 1 – 7 & 23 – 33 (odds) Concept

Concept

For f(x) = –x2 – 2x – 2, a = –1, b = –2, and c = –2. Maximum and Minimum Values A. Consider f(x) = –x2 – 2x – 2. Determine whether the function has a maximum or a minimum value. For f(x) = –x2 – 2x – 2, a = –1, b = –2, and c = –2. Answer: Because a is negative the graph opens down, so the function has a maximum value. Example 4

The maximum value is the y-coordinate of the vertex. Maximum and Minimum Values B. Consider f(x) = –x2 – 2x – 2. State the maximum or minimum value of the function. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is or –1. f(x) = –x2 – 2x – 2 Original function f(–1) = –(–1)2 – 2(–1) – 2 x = –1 f(–1) = –1 Simplify. Answer: The maximum value is –1. Example 4

Maximum and Minimum Values C. Consider f(x) = –x2 – 2x – 2. State the domain and range of the function. Answer: The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y | y  –1}. Example 4

A. Consider f(x) = 2x2 – 4x + 8. Determine whether the function has a maximum or a minimum value. B. minimum C. neither 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 Example 4

B. Consider f(x) = 2x2 – 4x + 8. State the maximum or minimum value of the function. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 Example 4

A. Domain: all real numbers; Range: {y | y ≥ 6} C. Consider f(x) = 2x2 – 4x + 8. State the domain and range of the function. A. Domain: all real numbers; Range: {y | y ≥ 6} B. Domain: all positive numbers; Range: {y | y ≤ 6} C. Domain: all positive numbers; Range: {y | y ≥ 8} D. Domain: all real numbers; Range: {y | y ≤ 8} 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 Example 4

Concept

Graph the function f(x) = –x2 + 5x – 2. Graph Quadratic Functions Graph the function f(x) = –x2 + 5x – 2. Step 1 Find the equation of the axis of symmetry. Formula for the equation of the axis of symmetry a = –1 and b = 5 or 2.5 Simplify. Example 5

f(x) = –x2 + 5x – 2 Original equation Graph Quadratic Functions Step 2 Find the vertex, and determine whether it is a maximum or minimum. f(x) = –x2 + 5x – 2 Original equation = –(2.5)2 + 5(2.5) – 2 x = 2.5 = 4.25 Simplify. The vertex lies at (2.5, 4.25). Because a is negative the graph opens down, and the vertex is a maximum. Example 5

Step 3 Find the y-intercept. Graph Quadratic Functions Step 3 Find the y-intercept. f(x) = –x2 + 5x – 2 Original equation = –(0)2 + 5(0) – 2 x = 0 = –2 Simplify. The y-intercept is –2. Example 5

Graph Quadratic Functions Step 4 The axis of symmetry divides the parabola into two equal parts. So if there is a point on one side, there is a corresponding point on the other side that is the same distance from the axis of symmetry and has the same y-value. Example 5

Step 5 Connect the points with a smooth curve. Graph Quadratic Functions Step 5 Connect the points with a smooth curve. Answer: Example 5

Graph the function f(x) = x2 + 2x – 2. A. B. C. D. Example 5

Graph the height of the arrow. Use a Graph of a Quadratic Function A. ARCHERY Ben shoots an arrow. The path of the arrow can be modeled by y = –16x2 + 100x + 4, where y represents the height in feet of the arrow x seconds after it is shot into the air. Graph the height of the arrow. Equation of the axis of symmetry a = –16 and b = 100 Example 6

y = –16x2 + 100x + 4 Original equation Use a Graph of a Quadratic Function The equation of the axis of symmetry is x = . Thus, the x-coordinate for the vertex is . y = –16x2 + 100x + 4 Original equation Simplify. The vertex is at . Example 6

Use a Graph of a Quadratic Function Let’s find another point. Choose an x-value of 0 and substitute. Our new point is (0, 4). The point paired with it on the other side of the axis of symmetry is Example 6

Use a Graph of a Quadratic Function Repeat this and choose an x-value to get (1, 88) and its corresponding point Connect these with points and create a smooth curve. Answer: Example 6

At what height was the arrow shot? Use a Graph of a Quadratic Function B. ARCHERY Ben shoots an arrow. The path of the arrow can be modeled by y = –16x2 + 100x + 4, where y represents the height in feet of the arrow x seconds after it is shot in the air. At what height was the arrow shot? The arrow is shot when the time equals 0, or at the y-intercept. Answer: The arrow is shot when the time equal 0, or at the y-intercept. So, the arrow was 4 feet from the ground when it was shot. Example 6

What is the maximum height of the arrow? Use a Graph of a Quadratic Function C. ARCHERY Ben shoots an arrow. The path of the arrow can be modeled by y = –16x2 + 100x + 4, where y represents the height in feet of the arrow x seconds after it is shot in the air. What is the maximum height of the arrow? The maximum height of the arrow occurs at the vertex. Example 6

A. TENNIS Ellie hit a tennis ball into the air A. TENNIS Ellie hit a tennis ball into the air. The path of the ball can be modeled by y = –x2 + 8x + 2, where y represents the height in feet of the ball x seconds after it is hit into the air. Graph the path of the ball. A. B. C. D. Example 6

B. TENNIS Ellie hit a tennis ball into the air B. TENNIS Ellie hit a tennis ball into the air. The path of the ball can be modeled by y = –x2 + 8x + 2, where y represents the height in feet of the ball x seconds after it is hit into the air. At what height was the ball hit? A. 2 feet B. 3 feet C. 4 feet D. 5 feet 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 Example 6

C. TENNIS Ellie hit a tennis ball into the air C. TENNIS Ellie hit a tennis ball into the air. The path of the ball can be modeled by y = –x2 + 8x + 2, where y represents the height in feet of the ball x seconds after it is hit into the air. What is the maximum height of the ball? A. 5 feet B. 8 feet C. 18 feet D. 22 feet 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 Example 6

Assignment:1/6/16 Page 549 – 550 Problems 9 – 19 (odds) Concept